5.5A: \(BH_3\)

This module will focus on the how Molecular Orbital theory applied to BH3. First, we would go through the the structure, symmetry elements that BH3 has. Next, we would talk about the symmetry labels due to the orbitals of the B atoms in BH3. Lastly, we would use all of the information about symmetry labels in BH3 to construct a BH3 diagram.

Introduction

Molecular Orbital theory is used to show how bonds between atoms in a molecules are formed from the orbital perspective. This theory is very important in understanding whether a molecule is paramagnetic or diamagnetic since Valence Bond theory can not establish this. We usually see the MO diagram of diatomic such H2, F2 or HBr. With this module we would learn how make MO diagram for polyatomic molecules and particularly BH3. The first step of making MO diagram for BH3 is to know the structure and symmetry elements BH3 has.

Symmetry Elements

The molecular structure of BH3 is trigonal planar and it belongs to the point group D3h. The symmetry elements are included:

b. How to determine the shapes of the LGOS?

-LGO's symmetry labels are a1' + e'. There are 3 LGOs that can be made out of these symmetry labels,one LGO is from a1' and two LGOs are from e' due to doubly degenerate. In order to determine the shape of each LGO, we would use the wavefunctions.

-Three hydrogens in BH3 are assigned with Ψ1, Ψ2, Ψ3. Now lets look at how each Ψ is affected by the symmetry operations of the D3h and their results are completed in the following table:

D3h

E

C3

C23

C2

C2’

C2’’

σh

S3

S23

σv

σv’

σv’’

Ψ1

Ψ1

Ψ2

Ψ3

Ψ1

Ψ3

Ψ2

Ψ1

Ψ2

Ψ3

Ψ1

Ψ3

Ψ2

a1’

1

1

1

1

1

1

1

1

1

1

1

1

LGO1

Ψ1

Ψ2

Ψ3

Ψ1

Ψ3

Ψ2

Ψ1

Ψ2

Ψ3

Ψ1

Ψ3

Ψ2

Ψ (a1') = 4Ψ1+4Ψ2+4Ψ3

= 4(Ψ1+Ψ2+Ψ3)

Ψ(a1')= 1/√3 (Ψ1+Ψ2+Ψ3)

-The shape of the LGO1 is

D3h

E

C3

C23

C2

C2’

C2’’

σh

S3

S23

σv

σv’

σv’’

Ψ1

Ψ1

Ψ2

Ψ3

Ψ1

Ψ3

Ψ2

Ψ1

Ψ2

Ψ3

Ψ1

Ψ3

Ψ2

e’

2

-1

-1

0

0

0

2

-1

-1

0

0

0

LGO1

2Ψ1

-Ψ2

-Ψ3

0

0

0

2Ψ1

-Ψ2

-Ψ3

0

0

0

Ψ(e') = 4 (Ψ1)-2 (Ψ2)-2 (Ψ3)

= 2[ 2(Ψ1)-Ψ2-Ψ3]

Ψ(e') = 1/ √6 (2 Ψ1-Ψ2-Ψ3)

-The shape of the LGO2 is

-Noticed that in the LGO2, we have 1 nodal plane which is the horizonal line between the positive charge and negative charge. Therefore, the LGO3 (doubly degenerate with e') would also 1 nodal plane and its wavefunction would be Ψ(e') = 1/√2 (Ψ2-Ψ3). The shape of the LGO3 is

MO diagram

-As we can see in this diagram, the energy level of 3 LGOs are higher than the 2s orbital and below the 2 p orbital dued to the electronegativy of both Boron and Hydrogen. Hydrogen has higher electronegativity than boron, therefore hydrogen would have lower energy level in the MO diagram.

-In addition, B has 3 electrons in the valence electrons and 3 hydrogens have total 3 electrons. Therefore, the total number of electrons filled in orbitals are 6. With all of the informations above about symmetry labels of B atom and the 3 LGOs, we now construct the MO diagram of BH3. Noticed that, the bonding formation only happens to atoms that have the same symmetry labels. 2s orbital and LGO(1) would contribute 1 electron to give 2 spin pairs electrons at the a1' energy level. 2px and 2py orbitals would bond to the LGO(2) and LGO(3), which give 2 spin pairs electrons at the e' energy level.